Given public generator $g$ of some cyclic group, a secrets $x\in Z_q$, and public pairs $(a_1,b_1),...,(a_n,b_n)$ (where $a_1,...,a_n$ are selected at random from a big set), and prime p, that satisfies the equations $g^{x^{a_i}}\equiv b_i\ (mod\ p)$ for all $i=1,...,n$.
Is it hard to compute to compute x (assuming the hardness of DLOG)?
And a weaker version of the above problem, given a new group element $b_{i+1}$, is it hard to compute an integer $a_{i+1}$ such that $g^{x^{a_{i+1}}}=b_{i+1}$?
A little notation: we'll write $\mathrm{DLOG}(h,y)$ for a magic, general purpose discrete logarithm solver that returns $z$ where $h^z=y$ in our group of order $q$ and $\mathrm{DH}(h,c,d)$ for a magic, general purpose Diffie-Hellman solver in our group which returns $h^{ef}$ where $c=h^e$ and $d=h^f$.
For your first question, if we have a magic DLOG solver then your problem is not hard. If we call our solver with $h=b_i$ and $y=b_j$ we will get the answer $x^{a_j-a_i}\mod q$. If any of pair of $a_i$s differ by 1 this immediately gives us $x$, otherwise pick two pairs $(i,j)$ and $(k,\ell)$ such that $(a_j-a_i)$ and $(a_\ell-a_k)$ are coprime (it is overwhelmingly likely that such pairs exist). Now by Bezout's identity there are values $r$ and $s$ such that $r(a_j-a_i)+s(a_\ell-a_k)=1$ and we see that $$\mathrm{DLOG}(b_i,b_j)^r\mathrm{DLOG}(b_k,b_\ell)^s\equiv x\pmod q.$$
On the other hand, if your an easy solution to your first problem exists this would provide an easy proof that Diffie-Hellman is exactly as hard as the discrete logarithm problem. This is an open conjecture and so we do not know how to demonstrate the reverse implication. To see how this would show that $\mathrm{DH}\rightarrow \mathrm{DLOG}$ note that a Diffie-Hellman oracle can be used to create additional $(a,b)$ instances by $\mathrm{DH}(g,b_i,b_j)=g^{x^{a_i+a_j}}$ and starting from $g^{x^1}$ we can use our DH oracle to compute $b$ for any given $a$.
